Compute the first three terms of the Taylor series about $x = 0$ of $\sqrt{1+x}$

Question

I'm taking 'Single Variable Calculus' and asked : Compute the first three terms of the Taylor series about $x = 0$ of $\sqrt{1+x}$

How do I go about solving this ? I'm not asking for an explicit solution but I'm not familiar with the term

about $x = 0$

I do not think what is being asked for is compute $x=0,x=1,x=2$ for the term $\sqrt{1+x}$ ?

Update :

How does $\sqrt{1+x}$ relate to Taylor series ? Is 'about x = 0' a math term in relation to Taylor series ?

Answer 1

When it says about x=0, it means that $a=0$ in $f(a)+f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 ...$. And no it is not asking for computing $x=0, x=1, x=2$, but rather the first three terms in the Taylor series, so $f(a)+f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2$ and disregarding the rest.

Responding to the update:

As it says about x=0, we shall substitute a=0 into the formula I have above.

If $f(x) = \sqrt{1+x}$, then $f(0) = 1$. Thus our first term is $1$. Now for our second term. $f'(x) = \frac{1}{2\sqrt{1+x}}$, so $f'(0) = \frac{1}{2}$. Thus our second term is $\frac{x}{2}$. Our third term is calculated the same way: $f''(x) = \frac{-1}{4(1+x)\sqrt{1+x}} \rightarrow f''(0) = \frac{-1}{4}$, thus the third term is $\frac{-x^2}{4*2!} = \frac{-x^2}{8}$. So the first three terms will be:

$$1+\frac{x}{2}+\frac{-x^2}{8}$$

Answer 2

$$f(x)=f(a)+\frac{f^{'}(a)}{1!}(x-a)+\frac{f^{''}(a)}{2!}(x-a)^2 ...= 1+\frac{1}{2}x-\frac{1}{8}x^2 ...$$